README_EN.md

53. Maximum Subarray

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Description

Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

 

Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.

Example 2:

Input: nums = [1]
Output: 1

Example 3:

Input: nums = [5,4,-1,7,8]
Output: 23

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -105 <= nums[i] <= 105

 

Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

Solutions

1. Dynamic programming

2. Divide and Conquer

Python3

Dynamic programming.

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        res = f = nums[0]
        for num in nums[1:]:
            f = num + max(f, 0)
            res = max(res, f)
        return res

Divide and Conquer.

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        def crossMaxSub(nums, left, mid, right):
            lsum = rsum = 0
            lmx = rmx = float('-inf')
            for i in range(mid, left - 1, -1):
                lsum += nums[i]
                lmx = max(lmx, lsum)
            for i in range(mid + 1, right + 1):
                rsum += nums[i]
                rmx = max(rmx, rsum)
            return lmx + rmx

        def maxSub(nums, left, right):
            if left == right:
                return nums[left]
            mid = (left + right) >> 1
            lsum = maxSub(nums, left, mid)
            rsum = maxSub(nums, mid + 1, right)
            csum = crossMaxSub(nums, left, mid, right)
            return max(lsum, rsum, csum)

        left, right = 0, len(nums) - 1
        return maxSub(nums, left, right)

Java

Dynamic programming.

class Solution {
    public int maxSubArray(int[] nums) {
        int f = nums[0], res = nums[0];
        for (int i = 1, n = nums.length; i < n; ++i) {
            f = nums[i] + Math.max(f, 0);
            res = Math.max(res, f);
        }
        return res;
    }
}

Divide and Conquer.

class Solution {
    public int maxSubArray(int[] nums) {
        return maxSub(nums, 0, nums.length - 1);
    }

    private int maxSub(int[] nums, int left, int right) {
        if (left == right) {
            return nums[left];
        }
        int mid = (left + right) >>> 1;
        int lsum = maxSub(nums, left, mid);
        int rsum = maxSub(nums, mid + 1, right);
        return Math.max(Math.max(lsum, rsum), crossMaxSub(nums, left, mid, right));
    }

    private int crossMaxSub(int[] nums, int left, int mid, int right) {
        int lsum = 0, rsum = 0;
        int lmx = Integer.MIN_VALUE, rmx = Integer.MIN_VALUE;
        for (int i = mid; i >= left; --i) {
            lsum += nums[i];
            lmx = Math.max(lmx, lsum);
        }
        for (int i = mid + 1; i <= right; ++i) {
            rsum += nums[i];
            rmx = Math.max(rmx, rsum);
        }
        return lmx + rmx;
    }
}

C++

class Solution {
public:
    int maxSubArray(vector<int>& nums) {
        int f = nums[0], res = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            f = nums[i] + max(f, 0);
            res = max(res, f);
        }
        return res;
    }
};

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxSubArray = function (nums) {
  let f = nums[0],
    res = nums[0];
  for (let i = 1; i < nums.length; ++i) {
    f = nums[i] + Math.max(f, 0);
    res = Math.max(res, f);
  }
  return res;
};

Go

func maxSubArray(nums []int) int {
    f, res := nums[0], nums[0]
    for i := 1; i < len(nums); i++ {
        if f > 0 {
            f += nums[i]
        } else {
            f = nums[i]
        }
        if f > res {
            res = f
        }
    }
    return res
}

C#

public class Solution {
    public int MaxSubArray(int[] nums) {
        int res = nums[0], f = nums[0];
        for (int i = 1; i < nums.Length; ++i)
        {
            f = nums[i] + Math.Max(f, 0);
            res = Math.Max(res, f);
        }
        return res;
    }
}

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